PHY2049: Chapter 221Chapter 22: Electric FieldPHY2049: Chapter 222Concept QuizÎFor which situations can we put a charge q’ on the leftside of the two charges so that it is in equilibrium? (1) (a) only (2) (a) and (b) (3) (c) and (d) (4) (b) and (d) (5) (c) onlyq′q′q′q′Should be near the smaller chargebecause if it is near the larger charge,there can be no total cancellationPHY2049: Chapter 223Concept QuizÎWhich situation has the largest force acting on q?qqqqAll E fields are in the same direction in (3). Note that (4) has E = 0.PHY2049: Chapter 224Electric Field of Single Point Charge2ˆkqErr=G2ˆkqErr=G0q >0q <PHY2049: Chapter 225Example: Electric Field on ProtonÎAt surface of proton q = e = 1.6 x 10-19C r = 10-15mÎE points radially outward for + charge()()()919212215910 1.6101.44 10 N/C10kqEr−−××== = ×PHY2049: Chapter 226E Field of Two Equal, Positive Point ChargesPHY2049: Chapter 227E Field of Two Equal, Unlike Point ChargesPHY2049: Chapter 228Field Between Two Charged Parallel PlatesÎAssume plates are much larger than separation E is approx. constant between plates E is zero outside the plates This is a capacitor!ÎE points from + plate to – plateÎWe will calculate E in Chap. 23 Gauss’ law Proportional to surface charge densityPHY2049: Chapter 2291. Rank magnitude of E at P1, P2, P3.Assume charges on rings are +q and +qPHY2049: Chapter 2210Answer to Question #1ÎP1has E = 0 (equidistant from ring A and B and they are same sign)ÎP3has largest E (contributions from ring A and B)ÎP2has no contribution from ring B because it is at the center, thus it is only affected by ring A.ÎSo the order (smallest E to largest E) is P1, P2, P3PHY2049: Chapter 22112. Which point has largest E?Assume charges on rings are +q and −qPHY2049: Chapter 2212Answer to Question #2ÎP1has largest E field since it is equidistant from ring A and B and their E contributions add, rather than cancel, as in the first question.PHY2049: Chapter 2213Calculate E of Dipole (⊥ axis)ÎAt point x, Ex= 0 and Ey< 0. Why?+Q-Qd()222 3/222222/2sin 2/4/4/4ykQ kQ d kQdErxdxdxdθ−− −⎛⎞== =⎜⎟+⎝⎠++3ykpExdx≈− (dipole moment)pQd=xrr22/4rxd=+/2sindrθ=θPHY2049: Chapter 2214Calculate E of Dipole (along axis)ÎAt point x, Ex> 0 and Ey= 0. Why?(dipole moment)pQd=x−Qd()()()22 2222/2 /2/4xkQ kQ kQxdExd xdxd=−=−+−32xkpExdx≈ x+QShow this yourself(algebra)PHY2049: Chapter 2215Finding E Field from Charge DistributionÎPerform integral over charge distribution Each component must be calculated separately (vector addition)ÎGeneral helpful rules Use symmetry to see if any component must be zero Use symmetry to see if any component is doubled, etc. Express dq, r and trig functions in terms of “natural” variables defined by the problem Then we can integrate!()2sin or cosykdqdErθθ=PHY2049: Chapter 2216Find E at Center of Uniformly Charged Circleconstr=dq ds rdλλθ==+q on top half−q on the bottom halfSymmetry:Ex= 0: Ex(right) = −Ex(left)Eydoubled: Ey(top) = +Ey(bottom)charge / unit lengthqrλπ==PHY2049: Chapter 2217Center of Uniformly Charged CircleÎFind E Express dq, r, sinθ in terms of θ Top, bottom give same contribution()22sin sinykrdkdqdErrλθθθ=− =−()2002sin2cos4ykrdErkrkrππλθθλθλ=× −−=−−=∫qrλπ=24ykqErπ=−dq ds
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